jq：如何在嵌套数组中进行选择答案

【问题标题】：jq: How to select within a nested arrayjq：如何在嵌套数组中进行选择
【发布时间】：2020-08-31 21:41:22
【问题描述】：

我正在尝试在 jq 中选择对象子集的子集，同时保留外部对象。

以输入为例：

{
  "head": "master",
  "hash": "ba2054c38ec68b73926acb24cd0bad61177a03d4",
  "upstream": "origin/master",
  "ahead": 1,
  "behind": 0,
  "changes": [
    {
      "name": "exe/each-package",
      "status": "updated_in_index",
      "work_tree_status": "modified",
      "index_status": "unmodified"
    },
    {
      "name": "exe/find-package",
      "status": "not_updated",
      "work_tree_status": "unmodified",
      "index_status": "modified"
    },
    {
      "name": "lib/tools/tool_support.rb",
      "status": "not_updated",
      "work_tree_status": "unmodified",
      "index_status": "modified"
    },
    {
      "name": "curses.txt",
      "status": "untracked",
      "work_tree_status": "untracked",
      "index_status": "untracked"
    },
    {
      "name": "exe/fat-stat",
      "status": "untracked",
      "work_tree_status": "untracked",
      "index_status": "untracked"
    },
    {
      "name": "exe/prefix-each-with",
      "status": "untracked",
      "work_tree_status": "untracked",
      "index_status": "untracked"
    },
    {
      "name": "notes.txt",
      "status": "untracked",
      "work_tree_status": "untracked",
      "index_status": "untracked"
    }
  ]
}

以及我删除了changes 成员子集的预期输出：

{
  "head": "master",
  "hash": "ba2054c38ec68b73926acb24cd0bad61177a03d4",
  "upstream": "origin/master",
  "ahead": 1,
  "behind": 0,
  "changes": [
    {
      "name": "exe/each-package",
      "status": "updated_in_index",
      "work_tree_status": "modified",
      "index_status": "unmodified"
    },
    {
      "name": "exe/find-package",
      "status": "not_updated",
      "work_tree_status": "unmodified",
      "index_status": "modified"
    },
    {
      "name": "lib/tools/tool_support.rb",
      "status": "not_updated",
      "work_tree_status": "unmodified",
      "index_status": "modified"
    }
  ]
}

我能够做到以下几点。但我怀疑有一种更简洁的方法来过滤嵌套数组的内部结构，也许可以消除重复。

jq '. + {changes: [.changes[] | select(.status != "untracked")]}'

【问题讨论】：

标签： json select jq

【解决方案1】：

更简洁可以通过以下方式实现：

.changes |= map(select(.status != "untracked"))

【讨论】：

华丽！我喜欢它。